`2.56g` of sulfur in `100 g` of `CS_(2)` of has depression in freez point of `0.01^(@)C.K_(f)=0.1^(@)` mol `al^(-1)`. Hence, he atomicity of sulfur is `CS_(2)` is

2.56 g of suphur in 100 g of CS_(2) has depression in freezing point of 0.010^(@)C , K_(f)=0.1^(@)C (molal)^(-1) . Hence atomicity of sulphur in the solution is

2.56 f of suphur in 100 g of CS_(2) has depression in freezing point of 0.010^(@) K_(f)=0.1^(@) (molal)^(-1) . Hence atomicity of sulphurin the solution is

2.56 g of sulphur in 100 g of CS_2 has depression in freezing point of 0.010^@ . K_(f)=0.1^(@) kg (mol)^(-1) . Hence atomicity of sulphur in the solution is

When 2.56 g of sulphur was dissolved in 100 g of CS_(2) , the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_(x)) . ( K_(f) for CS_(2) = 3.83 K kg mol^(-1) , Atomic mass of sulphur = 32g mol^(-1) ]

When 2.56 g of sulphur is dissolved in 100 g of CS_(2) , the freezing point of the solution gets lowerd by 0.383 K. Calculate the formula of sulphur (S_(x)) . [Given K_(f) for CS_(2)=3.83 "K kg mol"^(-1) ], [Atomic mass of sulphur=32g mol^(-1) ]

(a) When 2.56 g of sulphur was dissolved in 100 g of CS_(2) , the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_(x)) . ( K_(f) for CS_(2) = 3.83 K kg mol^(-1) , Atomic mass of sulphur = 32g mol^(-1) ] (b) Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing. (i) 1.2% sodium chloride solution? (ii) 0.4% sodium chloride solution?

When 2.56g of sulphur was dissolved in 100 g of CS_2 , the freezing point lowered by 0.383K. Calculate the formula of sulphur (S_x) . (k_(f) for CS_2 = 3.83K kg mol^-1 , Atomic mass of sulphur = 32g mol^-1) .