`6.02xx10^(20)` molecules of urea are present in 100 ml of its solution. The concentration of solution is :

Options

A. `90.0 g con HNO_(3)`
B. `70.0 g conc. HNO_(3)`
C. `54.0 g conc HNO_(3)`
D. `45.0 g conc . HNO_(3)`

Answer

D

Explanation

Molarity `= ("Moles of " HNO_(3))/("Volume of solution") xx 1000`
` 2 = ("Mass"//63)/(250) xx 1000`
Mass of `HNO_(3) = (2 xx 250 xx 63)/(1000) = 3.15 g`
Mass of `HNO_(3)` required for 70% solution `= (31.5 xx 100)/(70)`
` = 45.0 g