Question

Elements A and B form two compounds B2A3 and B2A. 0.05moles of B2A3 weigh 9g and 0.10 mole of B2A weigh 10g. Find Atomic weight of A and B

The volume occupied by 1.8g of water vapour at 374°C and 1bar pressure will be =?

Answer 1

1)

GIVEN:

• Two compounds : B2A3 and B2A
• 0.05 mol B2A3= 9g
• 0.10 mol B2A= 10g

NEED TO FIND:

• Atomic weight of A and B

ANSWER:

Let Atomic mass of A = y and mass of B= x

B2A3:-

0.05 mol= 9g

=> 1mol= 9 ÷0.05 g =900÷5g

=> 1mol= 180g

Thus as in this compound 2B and 3A is present, we can say:

$\boxed{\red{2x+3y=180}}$----i

B2A:

0.10 mol=10g

1mol= 10÷0.10 g

=> 1mol= 100g

As in the compound 2B and 1A is present we can say;

$\boxed{\red{2x+y=100}}$----ii

Now by i-ii we get:

2x+3y-2x-y=180-100

=>2y=80

=>$\bold{\boxed{\pink{y=Mass\:of\:A=40amu}}}$ (ans)

Thus putting y=40 in ii we get,

$\bold{\boxed{\pink{x=Mass\:of\:B=30amu}}}$ (ans)

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2)

GIVEN:

• Mass of water =1.8g
• Temperature= 374°C
• Pressure =1bar

NEED TO FIND:

• Volume=?

ANSWER:

We know,

let n= no of moles

=> n=1.8÷180 mol

=>n=0.1 mol

T= Temperature

=>T= 374°C = 647K

R= 0.083 (ideal gas constant)

Now by ideal gas equation;

$PV = nRT \\ \implies \: V = \dfrac{nRT}{P} \\ \implies \: V = \dfrac{0.1 \times 0.083 \times 647}{1} \\ \implies \: Volume = 5.3701L$