Question

Calculate the quantity of sodium carbonate (anhydrous ) required to prepare $250$ml solution.

• A. $2.65$ gram
• B. $4.95$ gram
• C. $6.25$ gram
• D. None of these

Answer 1

Answer: (A)

$2.65$ gram

$m=\frac{n}{v}=\frac{m}{m\times V_1\left(lt\right)}$

$\Rightarrow x=\frac{m}{106\times 25\times {10}^{-3}}$

$\Rightarrow m=x\times 106\times 25\times {10}^{-3}$

$=2.650$gm