Question

If 50 g oleum sample rated as 118% is mixed with 18 g water, then the correct option is

• A. The resulting solution contains only 118 g pure $H_{2}S{O}_4$
• B. The resulting solution contains 9 g of water and 59 g $H_{2}S{O}_4$
• C. The resulting solution contains 68 g of pure $H_{2}S{O}_4$
• D. The resulting solution contains 18 g of water and 118 g $H_{2}S{O}_4$

Answer 1

The correct option is B The resulting solution contains 9 g of water and 59 g $H_{2}S{O}_4$

If % label of oleum is 118% then this implies that from 100g of oleum sampel we can produce 118g of $H_{2}S{O}_4$ on addition of water.


Thus, it can be concluded that
100g of oleum sample can give 118g of $H_{2}S{O}_4$ on addition of water
1 g of sample will give $\frac{118}{100}$ g of $H_{s}S{O}_4$ on addition of water

And 50g of sample will give $\frac{118}{100}\times 50$
= 59 g of $H_{2}S{O}_4$
And mass of $H_{2}O$ used = 59 - 50 = 9g

Hence, resulting solution contains 9g of water and 59g of $H_{2}S{O}_4$

So, the correct answer is option (b).