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Question
CBSEENCH12010768
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1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol-1 is

  • 1186

  • 84.3

  • 118.6

  • 11.86

Solution

B.

84.3

Given chemical equation
M2CO3 +2HCl → 2MCl + H2O + CO2
1 gm                           0.01186 mol
From the balanced chemical equation
1/M = 0.01186
M = 84.3 gm/mol


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Question
CBSEENCH12010616
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18 g glucose (C6H12O6 ) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is:

  • 76.0

  • 752.4

  • 759.0

  • 7.6

Solution

B.

752.4

Vapour pressure of water (p°) = 760 torr
Number of moles of glucose
WiredFaculty
Molar mass of water = 18 g/mol
Mass of water (given) = 178.2 g
Number of moles of water

equals space fraction numerator Mass space of space water over denominator Molar space mass space of space water end fraction space equals space fraction numerator 178.2 over denominator 18 space straight g end fraction space equals space 9.9 space mol
Total number of moles = (0.1 +9.9) = 10 moles
Now, the mole fraction of glucose in solution = Change in pressure with respect to initial pressure.
WiredFaculty
therefore, Vapour pressure of solution = (760-7.6)torr
=752.4 torr

Question
CBSEENCH12010875
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18 g of glucose (C6H12O6) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100o C is

  • 759.00 Torr

  • 7.60 Torr

  • 76.00 Torr

  • 752.40 Torr

Solution

C.

76.00 Torr

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Question
CBSEENCH12010633
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3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is:

  • 18 mg

  • 36 mg

  • 42 mg

  • 54 mg

Solution

A.

18 mg

The initial strength of acetic acid = 0.06N
Final strength = 0.042 N
Volume given = 50 mL
there Initial m moles of CH3COOH 
 = 0.06 x 50 = 3
Final m moles of CH3COOH 
 = 0.042 x 50 = 21
therefore, m moles of CH3COOH absorbed
 = 3-2.1
 = 0.9 m mol
Hence, mass of CH3COOH  absorbed per gram of charcoal
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Question
CBSEENCH12010932
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6.02×1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is

  • 0.001 M

  • 0.1 M

  • 0.02 M

  • 0.01 M

Solution

D.

0.01 M

Avogadros number NA = 6.02 x 1023  = 1 mole
∴ 6.02 x 1020 molecules 
= 0.001 mole in 100 mL (0.1 L solution)
∴ Molar concentration  = mol/volume in L
 =0.001/0.1
 = 0.01 M

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