The relation between molarity and mole fraction isXB =MMAM(MA~MB)+dWhere XB = mole fraction of BM= molarityMA = Molecular mass of AMB = Molecular mass of Bd= densityXB =MMAM(MA~MB)+d0 6=M ×20M(20 ~10) + 0 7Solve for MMolarity= 0 03 MolarThe relation between molality and mole fraction isXB =mMA1+mMAWhere XB = mole fraction of Bm= molalityMA = Molecular mass of AXB =mMA1+mMA0 6 =m×201+m×20Solve for mMolality =0 075 m

Solve question no.2

Solve question no.2 02. A solution ofA (MM = 20) and B (MM = 10), [Mole fraction Xa = 0.6] having density 0.7 gm/ml then molarity and' and molality of B in this solution will be (a) 30M,75m (b) 40M,75m (c) 30M,65m respectively. (d) 50M,55m

T h e r e l a t i o n b e t w e e n m o l a r i t y a n d m o l e f r a c t i o n i s X_{B} = \frac{M M_{A}}{M (M_{A} ~ M_{B}) + d} W h e r e X_{B} = m o l e f r a c t i o n o f B M = m o l a r i t y M_{A} = M o l e c u l a r m a s s o f A M_{B} = M o l e c u l a r m a s s o f B d = d e n s i t y X_{B} = \frac{M M_{A}}{M (M_{A} ~ M_{B}) + d} 0.6 = \frac{M \times 20}{M (20 ~ 10) + 0.7} S o l v e f o r M M o l a r i t y = 0.03 M o l a r T h e r e l a t i o n b e t w e e n m o l a l i t y a n d m o l e f r a c t i o n i s X_{B} = \frac{m M_{A}}{1 + m M_{A}} W h e r e X_{B} = m o l e f r a c t i o n o f B m = m o l a l i t y M_{A} = M o l e c u l a r m a s s o f A X_{B} = \frac{m M_{A}}{1 + m M_{A}} 0.6 = \frac{m \times 20}{1 + m \times 20} S o l v e f o r m M o l a l i t y = 0.075 m

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Solve question no.2 Solve question no.2 02. A solution ofA (MM = 20) and B (MM = 10), [Mole fraction Xa = 0.6] having density 0.7 gm/ml then molarity and' and molality of B in this solution will be (a) 30M,75m (b) 40M,75m (c) 30M,65m respectively. (d) 50M,55m

Solve question no.2

Solve question no.2 02. A solution ofA (MM = 20) and B (MM = 10), [Mole fraction Xa = 0.6] having density 0.7 gm/ml then molarity and' and molality of B in this solution will be (a) 30M,75m (b) 40M,75m (c) 30M,65m respectively. (d) 50M,55m