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1

### AIPMT 2010 Prelims

25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na+ and carbonate ions, CO$${}_3^{2 - }$$ are respectively (Molar mass of Na2CO3 = 106 g mol$$-$$1)
A
0.955 M and 1.910 M
B
1.910 M and 0.955 M
C
1.90 M and 1.910 M
D
0.477 M and 0.477 M

## Explanation

Given in the question that molar mass of Na2CO3 = 106 g

$$\therefore$$ Molarity of solution = $${{25.3 \times 1000} \over {106 \times 250}}$$

= 0.9547 M = 0.955 M

Na2CO3 $$\to$$ 2Na+ + $$CO_3^{2-}$$

[Na+] = 2[Na2CO3] = 2 $$\times$$ 0.955 = 1.910 M

[$$CO_3^{2-}$$] = [Na2CO3] = 0.955 M
2

### AIPMT 2010 Prelims

The number of atoms in 0.1 mol of a triatomic gas is (NA = 6.02 $$\times$$ 1023 mol$$-$$1)
A
6.026 $$\times$$ 1022
B
1.806 $$\times$$ 1023
C
3.600 $$\times$$ 1023
D
1.800 $$\times$$ 1022

## Explanation

No. of atoms

= NA $$\times$$ No. of moles $$\times$$ 3

= 6.023 $$\times$$ 1023 $$\times$$ 0.1 $$\times$$ 3

= 1.806 $$\times$$ 1023
3

### AIPMT 2009

10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be
A
3 mol
B
4 mol
C
1 mol
D
2 mol

## Explanation

H2 + 1/2O2 $$\to$$ H2O
2 g 16 g 18 g
1 mol 0.5 mol 1 mol
10 g of H2 = 5 mol and 64 g og O2 = 2 mol

$$\therefore$$ In this reaction, oxygen is the limiting reagent so amount of H2O produced depends on that of O2

Since 0.5 mol of O2 gives 1 mol H2O

$$\therefore$$ 2 mol of O2 will give 4 mol H2O
4

### AIPMT 2008

What volume of oxygen gas (O2) measuread at 0oC and 1 atm, is needed to burn completely 1 L of propane gas (C3H8) measured under the same conditions ?
A
5 L
B
10 L
C
7 L
D
6 L

## Explanation

C3H8 + 5O2 $$\to$$ 3CO2 + 4H2O (balanced equation)
1 vol. 5 vol. 3 vol. 4 vol.

According to the above equation 1 vol. or 1 litre of propane requires to 5 vol. or 5 litre of O2 to burn completely.

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