Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is .

Options

A. 0.004
B. 0.222
C. 0.111
D. 0.021

Answer

C

Explanation

Relative lowering in vapoure pressure , `(p^(@) - p)/(p^(@)) = x_(2) = 0.002`
Molality of solution means the number of moles of glucose present in 100 g of water .
`w_(1) = 1000 g, n_(1) = (1000)/(18) = 55.55`
`(n_(2))/(n_(2) + 55.55) = 0.002`
For dilute solution , 55.55 + `n_(2) = 55.55`
`therefore n_(2) = 0.002 xx 55.55 = 0.1111`