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1

### AIPMT 2015

The number of water molecules is maximum in
A
1.8 gram of water
B
18 gram of water
C
18 moles of water
D
18 molecules of water

## Explanation

1.8 g of water = $${{6.023 \times {{10}^{23}}} \over {18}} \times 1.8$$

= 6.023 $$\times$$ 1022 molecules

$$\therefore$$ 18 g of water = 6.023 $$\times$$ 1023 molecules = 1 mole of water

$$\therefore$$ 18 moles of water = 18 $$\times$$ 6.023 $$\times$$ 1023 molecules

2

### AIPMT 2015

If Avogadro number NA, is changed from 6.022 $$\times$$ 1023 mol$$-$$1 to 6.022 $$\times$$ 1020 mol$$-$$1, this would change
A
the mass of one mole of carbon
B
the ratio of chemical species to each other in a balanced equation
C
the ratio of elements to each other in a compound
D
the definition of mass in units of grams.

## Explanation

We know mass of 1 mol (6.022 $$\times$$ 1023) atoms of carbon = 12 g

If Avogadro number is changed to 6.022 $$\times$$ 1020 then mass of 1 mol of cabon is

= $${{12 \times 6.022 \times {{10}^{20}}} \over {6.022 \times {{10}^{23}}}}$$ = $$12 \times {10^{ - 3}}\,g$$

$$\therefore$$ It would change the mass of one mole of carbon.
3

### AIPMT 2015

What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8% NaCl solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5
A
3.5 g
B
7 g
C
14 g
D
28 g

## Explanation

50 ml of 16.9% solution of AgNO3

$$\left( {{{16.9} \over {100}} \times 50} \right)$$ = 8.45 g of AgNO3

nmole = $${{8.45g} \over {(107.8 + 14 + 16 \times 3)g/mol}}$$

= $$\left( {{{8.45g} \over {169.8g/mol}}} \right) = 0.0497\,moles$$

50ml of 5.8% solution of NaCl contain

NaCl = $$\left( {{{5.8} \over {100}} \times 50} \right) = 2.9g$$

nNaCl = $${{2.9g} \over {(23 + 35.5)g/mol}}$$

= 0.0495 moles

AgNO3 + NaCl $$\Rightarrow$$ AgCl + Na$$\oplus$$ + Cl$$\ominus$$
1 mole 1 mole 1 mole
0.049 mole 0.049 mole 0.049 mole of AgCl

n = $${w \over M}$$

$$\Rightarrow$$ w = (nAgCl) $$\times$$ Molecular mass

= (0.049) $$\times$$ (107.8 + 35.5) = 7.02 g
4

### AIPMT 2014

1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel, Which reactant is left in excess and how much ? (At. wt. Mg = 24, O = 16)
A
Mg. 0.16 g
B
O2, 0.16 g
C
Mg, 0.44 g
D
O2 , 0.28 g

## Explanation

nMg = $$1 \over 24$$ = 0.0416 moles

nO2 = $$0.56 \over 32$$ = 0.0175 moles

Mg + $${1\over2} O_2$$ $$\Rightarrow$$ MgO
Initial 0.0416 moles 0.0175 moles 0
Final ( 0.0416 - 2 x 0.0175)
= 0.0066 moles
0 2 x 0.0175

$$\because$$ Mass of Mg left in excess = 0.0066 $$\times$$ 24 = 0.16 g

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