Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ_2. When 1 g of PQ is dissolved in 50 g of solvent ‘A’. Delta T_b was 1.176 K while when 1 g of PQ_2 is dissolved in 50 g of solvent ‘A’, Delta T_b was 0.689 K. (K_b of ‘A’ = 5text K kg mol^-1). The molar masses of elements P and Q (in textg mol^-1) respectively, are :

Solution & Explanation

### Related Formula Delta T_b = K_b times m m = fractextWeight of solute (g)textMolar mass of solute times frac1000textWeight of solvent (g) ### Core Logic For compound PQ: (Delta mathrmT_b)_mathrmPQ = mathrmK_b cdot m 1.176 = 5 times frac1mathrmM_1 times frac100050 mathrmM_1 = frac5 times 201.176 = 85.03 text g/mol For compound PQ_2: (Delta mathrmT_b)_mathrmPQ_2 = 5 times frac1mathrmM_2 times frac100050 = 0.689 mathrmM_2 = frac5 times 200.689 = 145.13 text g/mol Let molar mass of P & Q be mathrmM_P and mathrmM_Q respectively: mathrmM_P + mathrmM_Q = 85.03 quad text--- (1) mathrmM_P + 2mathrmM_Q = 145.13 quad text--- (2) Subtracting (1) from (2): mathrmM_Q = 145.13 - 85.03 = 60.1 approx 60 text g/mol Substituting back into (1): mathrmM_P + 60.1 = 85.03 implies mathrmM_P = 24.93 approx 25 text g/mol ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

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