Question

A chemist has m gram of salt water that is m% salts. How many grams of salt must he add to make the solution 2m% salt?
[a] $\dfrac{m}{100+m}$
[b] $\dfrac{2m}{100-2m}$
[c] $\dfrac{{{m}^{2}}}{100-2m}$
[d] $\dfrac{{{m}^{2}}}{100+2m}$
[e] $\dfrac{2m}{100+m}$

Answer 1

Hint: Assume that the amount of salt to be added be x. Find the weight of the solute in the solution using the fact that if the weight of the solution is ${{w}_{s}}$ and the weight of the solution is $w$ then the concentration c of the solution is given by $c=\dfrac{w}{{{w}_{s}}}\times 100$. Hence find the concentration of the final solution in terms of x. Equate this expression to 2m% and hence form an equation in x. Solve for x and hence find the amount of salt to be added.

Complete step by step answer:
Let the amount of salt to be added be x.
Now, we know that $c=\dfrac{w}{{{w}_{s}}}\times 100$, where c is the concentration , w is the weight of the solute and ${{w}_{s}}$v is the volume of the solution.
Now, since we have m gram of solution of m% concentrated solution, we have
$m=\dfrac{w}{m}\times 100$
Hence, we have $w=\dfrac{{{m}^{2}}}{100}$
Now when we add x gram of the solute, the total amount of solute in the solution $=\dfrac{{{m}^{2}}}{100}+x$
Also, the total weight of the solution becomes $m+x$
Hence, we have
$c=\dfrac{\dfrac{{{m}^{2}}}{100}+x}{m+x}\times 100$, where c is the concentration of the final solution.
But it is given that the concentration of the final solution is 2m%.
Hence, we have
$\begin{align}
  & 2m=\dfrac{\dfrac{{{m}^{2}}}{100}+x}{m+x}\times 100 \\
 & \Rightarrow 2m=\dfrac{{{m}^{2}}+100x}{m+x}\text{ }\left( i \right) \\
\end{align}$
Multiplying both sides by m+x, we get
$\begin{align}
  & \left( m+x \right)\left( 2m \right)={{m}^{2}}+100x \\
 & \Rightarrow 2{{m}^{2}}+2mx={{m}^{2}}+100x \\
\end{align}$
Subtracting 2mx from both sides, we get
${{m}^{2}}+x\left( 100-2m \right)=2{{m}^{2}}$
Subtracting ${{m}^{2}}$ from both sides, we get
$x\left( 100-2m \right)={{m}^{2}}$
Dividing both sides by 100 – 2m, we get
$x=\dfrac{{{m}^{2}}}{100-2m}$
Hence option [c] is correct.

Note: Alternatively, we can solve equation (i) directly using componendo-dividendo which states that if $\dfrac{a}{b}=\dfrac{c}{d}$ then $\dfrac{{{k}_{1}}a+{{k}_{2}}b}{{{k}_{3}}a+{{k}_{4}}b}=\dfrac{{{k}_{1}}c+{{k}_{2}}d}{{{k}_{3}}c+{{k}_{4}}d}$
Hence, we have
$\begin{align}
  & \dfrac{2m-m\left( 1 \right)}{2m-100\left( 1 \right)}=\dfrac{{{m}^{2}}+100x-m\left( m+x \right)}{{{m}^{2}}+100x-100\left( m+x \right)} \\
 & \Rightarrow \dfrac{m}{2m-100}=\dfrac{x\left( 100-m \right)}{m\left( m-100 \right)} \\
 & \Rightarrow x=\dfrac{{{m}^{2}}}{100-2m} \\
\end{align}$
Which is the same as obtained above.
Hence option [c] is correct.