The freezing point of solution made by dissolving 1.1 g of `CoCl_2.6NH_3` (molecular weight = 267) in 100 g `H_2O is - 0.29^@C`. How many moles of solute particles exist in solution for each mole of solute introduced? `(K_b= 1.86^@C m^(-1))`

Options

A. 2
B. 3
C. 4
D. 1

Answer

C

Explanation

Molality (experimental) = rimental) `(DeltaT_f)/(K_f)=(0.29)/(1.86)`
molality (theoretical) `= (1.1)/267 xx 1000/100 =11/267`
Thus, number of solute particles produced by 1 mole of solute i.e.,
`I = (0.29)/(1.86)xx 207/11 = 4`