Question

A chemist has $m$ gm of salt water that is $m$% salts. How many gram of salt must he add to make a solution that is $2m$% salt?

• A. $\frac{m}{100+m}$
• B. $\frac{2m}{100-2m}$
• C. $\frac{m^2}{100-2m}$
• D. $\frac{m^2}{100+2m}$
• E. $\frac{2m}{100+m}$

Answer 1

Answer: (C)

$\frac{m^2}{100-2m}$

$m$ gm of salt water

$m\%$ of salt in water

$\Rightarrow m\times \frac{m}{100}$ gram of salt

$\Rightarrow m-\frac{m^2}{100}$gram of water

Now, for $2m\%$ salt solution,

$\Rightarrow M\times \frac{2m}{100}$gram of salt in water($M$ is changed total mass)

$\Rightarrow M-\frac{M\left(2m\right)}{100}$ gram of water which is equal to the earlier one,

$\Rightarrow M(1-\frac{2m}{100})=(m-\frac{m^2}{100})$

$\Rightarrow M=\frac{100m-m^2}{100-2m}$ - putting it to grams of salt

$\Rightarrow$ gram of salt$=\frac{m(100-m)}{100-2m}\times \frac{2m}{100}=\frac{m^2}{100-2m}$