The freezing point of solution containing 10 mL of non-volatile and non-electrolyte liquid "A" in 500 g of water is `-0.413^@C`. If `K_f` of water is 1.86 K kg `mol^(-1)` and the molecular weight of `A=60 g mol^(-1)` , what is the density of the solution in g `mL^(-1)` ?
(Assume `Delta_(mix)` V=0)

Options

A. 1.13
B. 1.3
C. 0.9
D. 0.993

Answer

D

Explanation

Given , Freezing point of solution `=-0.413^@C`
Volume of non-volatile and non electrolyte liquid , A i.e. , solute = 10 mL
Weight of water i.e., solvent , `W_S`=500 g
`K_f` of water = 1.86 k kg `mol^(-1)`
Molecular weight of A , M = 60 g `mol^(-1)`
`Delta_(mix)V=0`
Density of solution,
`d="Total mass"/"Total volume" = (W+W_S)/(V+V_A)` ...(1)
Since, `DeltaT_f = K_f W/M xx 1000 / W_S`
`0-(-0.413)=K_f W/M xx 1000/W_S=1.86xxW/60xx1000/500`
`rArr` 0.413=0.062 W
`rAr W=0.413/0.062`
W=6.66 g
Then equation (1) becomes ,
`d=(6.66+500)/(500+10)`
`=506.66/510`
`therefore d=0.993 gmL^(-1)`