5 moles of AB2 weigh 125- 10-3kg and 10 moles of A2B2 weigh 300- 10-3kg-The molar mass of AMA and molar mass of BMB in kg-mol are -

Number of moles=Given massMolar mass For nbsp;AB_2 nbsp; 5=125M_A+2M_B M_A+2M_B=25..........(1) For nbsp;A_2B_2 10=3002M_A+2M_B 2M_A+2M_B=30..........(2) ...

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Standard XII

Chemistry

Stoichiometric Calculations

5 moles of AB...

Question

5 moles of $A B_{2}$ weigh $125 \times 10^{- 3} k g$ and 10 moles of $A_{2} B_{2}$ weigh $300 \times 10^{- 3} k g$ .The molar mass of $A (M_{A})$ and molar mass of $B (M_{B})$ in kg/mol are :

M_{A} = 10 \times 10^{- 3} a n d M_{B} = 5 \times 10^{- 3}

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M_{A} = 25 \times 10^{- 3} a n d M_{B} = 50 \times 10^{- 3}

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M_{A} = 5 \times 10^{- 3} a n d M_{B} = 10 \times 10^{- 3}

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M_{A} = 50 \times 10^{- 3} a n d M_{B} = 25 \times 10^{- 3}

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Solution

The correct option is C $M_{A} = 5 \times 10^{- 3} a n d M_{B} = 10 \times 10^{- 3}$
$Number of moles = \frac{Given mass}{Molar mass}$
For $A B_{2}$
$5 = \frac{125}{M_{A} + 2 M_{B}}$
$M_{A} + 2 M_{B} = 25$ ..........(1)
For $A_{2} B_{2}$
$10 = \frac{300}{2 M_{A} + 2 M_{B}}$
$2 M_{A} + 2 M_{B} = 30$ ..........(2)
Solving 1 and 2 ;
$M_{A} = 5 g / m o l = 5 \times 10^{- 3}$ kg/mol
$M_{B} = 10 g / m o l = 10 \times 10^{- 3}$ kg/mol

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5 moles of AB2 weigh 125×10−3kg and 10 moles of A2B2 weigh 300×10−3kg.The molar mass of A(MA) and molar mass of B(MB) in kg/mol are :

The correct option is C MA=5×10−3andMB=10×10−3Number of moles=Given massMolar mass For AB2 5=125MA+2MB MA+2MB=25..........(1) For A2B2 10=3002MA+2MB 2MA+2MB=30..........(2) Solving 1 and 2 ; MA=5g/mol=5×10−3kg/mol MB=10g/mol=10×10−3kg/mol

5 moles of $A B_{2}$ weigh $125 \times 10^{- 3} k g$ and 10 moles of $A_{2} B_{2}$ weigh $300 \times 10^{- 3} k g$ .The molar mass of $A (M_{A})$ and molar mass of $B (M_{B})$ in kg/mol are :