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But we always measure our weight using the unit kg! Are we right in doing so?

No. You are right, we are wrong in doing so.

Actually, what is wrong with it, is our use of the word weight. If you instead of saying "My weight is 70 kg" said "My mass is 70 kg", then everything is fine.

When you stand on a scale, the scale measures the force you exert, which is your weight $W$ in $\mathrm{[N]}$, and then by itself multiplies with $g$ to end up showing you your mass $m$ on the screen. It simply uses the equation:

$$W=mg$$

to find $m$. The calibration of the scale is the reason that the $g$ fits. If you brought a scale made on Earth to the Moon, it would be wrong! Since gravity is about 6 times weaker, which means that $g$ at the moon is 6 times smaller and you now weigh 6 times less, then (since the scale still uses the Earth's $g$) the mass (which should be constant nomatter where you are) that the scale shows on the screen will be 6 times too small as well:

$$W_{moon}=mg_{Earth} \implies \frac{1}{6}W_{Earth}=mg_{Earth} \implies \frac{1}{6}\frac{W_{Earth}}{g_{Earth}}=m$$

What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a gravitaional acceleration with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$

But we always measure our weight using the unit kg! Are we right in doing so?

No. You are correct in thinking that we are wrong in doing so.

Actually, what is wrong with it, is our use of the word weight. If you instead of saying "My weight is 70 kg" said "My mass is 70 kg", then everything is fine.

When you stand on a scale, the scale measures the force you exert, which is your weight $W$ in $\mathrm{[N]}$, and then by itself multiplies with $g$ to end up showing you your mass $m$ on the screen. It simply uses the equation:

$$W=mg$$

to find $m$. The calibration of the scale is the reason that the $g$ fits. If you brought a scale made on Earth to the Moon, it would be wrong! Since gravity is about 6 times weaker, which means that $g$ at the moon is 6 times smaller and you now weigh 6 times less, then (since the scale still uses the Earth's $g$) the mass (which should be constant nomatter where you are) that the scale shows on the screen will be 6 times too small as well:

$$W_{moon}=mg_{Earth} \implies \frac{1}{6}W_{Earth}=mg_{Earth} \implies \frac{1}{6}\frac{W_{Earth}}{g_{Earth}}=m$$

What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a gravitaional acceleration with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$

But we always measure our weight using the unit kg! Are we right in doing so?

No. You are right, we are wrong in doing so.

Actually, what is wrong with it, is our use of the word weight. If you instead of saying "My weight is 70 kg" said "My mass is 70 kg", then everything is fine.

When you stand on a scale, the scale measures the force you exert, which is your weight $W$ in $\mathrm{[N]}$, and then by itself multiplies with $g$ to end up showing you your mass $m$ on the screen. It simply uses the equation:

$$W=mg$$

to find $m$. The calibration of the scale is the reason that the $g$ fits. If you brought a scale made on Earth to the Moon, it would be wrong! Since gravity is about 6 times weaker, which means that $g$ at the moon is 6 times smaller and you now weigh 6 times less, then (since the scale still uses the Earth's $g$) the mass shown will be wrong.

What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a gravitaional acceleration with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$

But we always measure our weight using the unit kg! Are we right in doing so?

No. You are right, we are wrong in doing so.

Actually, what is wrong with it, is our use of the word weight. If you instead of saying "My weight is 70 kg" said "My mass is 70 kg", then everything is fine.

When you stand on a scale, the scale measures the force you exert, which is your weight $W$ in $\mathrm{[N]}$, and then by itself multiplies with $g$ to end up showing you your mass $m$ on the screen. It simply uses the equation:

$$W=mg$$

to find $m$. The calibration of the scale is the reason that the $g$ fits. If you brought a scale made on Earth to the Moon, it would be wrong! Since gravity is about 6 times weaker, which means that $g$ at the moon is 6 times smaller and you now weigh 6 times less, then (since the scale still uses the Earth's $g$) the mass (which should be constant nomatter where you are) that the scale shows on the screen will be 6 times too small as well:

$$W_{moon}=mg_{Earth} \implies \frac{1}{6}W_{Earth}=mg_{Earth} \implies \frac{1}{6}\frac{W_{Earth}}{g_{Earth}}=m$$

What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a gravitaional acceleration with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$

But we always measure our weight using the unit kg! Are we right in doing so?

No. You are right, we are wrong in doing so.

Actually, what is wrong with it, is our use of the word weight. If you instead of saying "My weight is 70 kg" said "My mass is 70 kg", then everything is fine.

What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a gravitaional acceleration with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$

But we always measure our weight using the unit kg! Are we right in doing so?

No. You are right, we are wrong in doing so.

Actually, what is wrong with it, is our use of the word weight. If you instead of saying "My weight is 70 kg" said "My mass is 70 kg", then everything is fine.

When you stand on a scale, the scale measures the force you exert, which is your weight $W$ in $\mathrm{[N]}$, and then by itself multiplies with $g$ to end up showing you your mass $m$ on the screen. It simply uses the equation:

$$W=mg$$

to find $m$. The calibration of the scale is the reason that the $g$ fits. If you brought a scale made on Earth to the Moon, it would be wrong! Since gravity is about 6 times weaker, which means that $g$ at the moon is 6 times smaller and you now weigh 6 times less, then (since the scale still uses the Earth's $g$) the mass shown will be wrong.

What will be the units of Newton? Will that be $N=kg \times g$?

No, you forgot to put the units of the "gravity" $g$ you mention in instead of $g$. Since $g$ that you call "gravity" is a gravitaional acceleration with units of $[m/s^2]$, the right unit equivalence is:

$$[N]=[kg] \cdot [m/s^2]$$