5 moles of AB2 weight 125×10−3 kg and 10 moles of A2B2 weight 300×10−3 kg. The molar mass of A(MA) and molar mass of B(MB) in kgmol−1 are

Key Idea To find the mass of A and B in the given question, mole concept is used. Number of moles (n)= molecular mass (M) given mass (w) Compound AB2A2B2 Mass of A(g)MA2MA Mass of B(g)2MB2MB We know that, Number of moles (n)= molecular mass (M) given mass (w) n×M=w…(A) Using equation (A), it can be concluded that 5(MA+2MB)=125×10−3 kg 10(2MA+2MB)=300×10−3 kg From equation (i) and (ii) 21(2MA+2MB)(MA+2MB)=(300125) On solving the equation, we obtain MA=5×10−3 and MB=10×10−3 So, the molar mass of A(MA) is 5×10−3 kg mol−1 and B(MB) is 10×10−3 kg mol−1.

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Some Basic Concepts of Chemistry

Mole concept

5 moles of A B_{2} weight 125 10^{-3} {~kg} and 10 moles of A

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5 moles of $A B_{2}$ weight $125 \times 1 0^{- 3} kg$ and 10 moles of $A_{2} B_{2}$ weight $300 \times 1 0^{- 3} kg$ . The molar mass of $A (M_{A})$ and molar mass of $B (M_{B})$ in $kg mol^{- 1}$ are

M_{A} = 10 \times 1 0^{- 3}

and

M_{B} = 5 \times 1 0^{- 3}

M_{A} = 50 \times 1 0^{- 3}

and

M_{B} = 25 \times 1 0^{- 3}

M_{A} = 25 \times 1 0^{- 3}

and

M_{B} = 50 \times 1 0^{- 3}

M_{A} = 5 \times 1 0^{- 3}

and

M_{B} = 10 \times 1 0^{- 3}

Views: 5,290 students

Updated on: Oct 4, 2023

Book:

Arihant Chemistry JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

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Harper Discussed

5 moles of

A B_{2}

weight

125 \times 1 0^{- 3} kg

and 10 moles of

A_{2} B_{2}

weight

300 \times 1 0^{- 3} kg

. The molar mass of

A (M_{A})

and molar mass of

B (M_{B})

kg mol^{- 1}

are

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

Key Idea To find the mass of

A

and

B

in the given question, mole concept is used. Number of moles

(n) = \frac{given mass ( w )}{molecular mass ( M )}

Compound A B_{2} A_{2} B_{2} Mass of A (g) M_{A} 2 M_{A} Mass of B (g) 2 M_{B} 2 M_{B}

We know that, Number of moles

(n) = \frac{given mass ( w )}{molecular mass ( M )}

n \times M = w \dots (A)

Using equation (A), it can be concluded that

5 (M_{A} + 2 M_{B}) = 125 \times 1 0^{- 3} kg

10 (2 M_{A} + 2 M_{B}) = 300 \times 1 0^{- 3} kg

From equation (i) and (ii)

\frac{1}{2} \frac{( M _{A} + 2 M _{B} )}{( 2 M _{A} + 2 M _{B} )} = (\frac{125}{300})

On solving the equation, we obtain

M_{A} = 5 \times 1 0^{- 3} and M_{B} = 10 \times 1 0^{- 3}

So, the molar mass of

A (M_{A})

5 \times 1 0^{- 3} kg mol^{- 1}

and

B (M_{B})

10 \times 1 0^{- 3} kg mol^{- 1}

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Question Text	5 moles of $A B_{2}$ weight $125 \times 1 0^{- 3} kg$ and 10 moles of $A_{2} B_{2}$ weight $300 \times 1 0^{- 3} kg$ . The molar mass of $A (M_{A})$ and molar mass of $B (M_{B})$ in $kg mol^{- 1}$ are
Updated On	Oct 4, 2023
Topic	Some Basic Concepts of Chemistry
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 20
Upvotes	2096
Avg. Video Duration	8 min

5 moles of $A B_{2}$ weight $125 \times 1 0^{- 3} kg$ and 10 moles of $A_{2} B_{2}$ weight $300 \times 1 0^{- 3} kg$ . The molar mass of $A (M_{A})$ and molar mass of $B (M_{B})$ in $kg mol^{- 1}$ are

Text solutionVerified

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5 moles of $A B_{2}$ weight $125 \times 1 0^{- 3} kg$ and 10 moles of $A_{2} B_{2}$ weight $300 \times 1 0^{- 3} kg$ . The molar mass of $A (M_{A})$ and molar mass of $B (M_{B})$ in $kg mol^{- 1}$ are