NEET
Practice Test 1
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A gas occupies a volume of 300 cc at 27°C and 620 mm pressure. The volume of gas at 47°C and 640 mm pressure is:

  • Solution

    From \(\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\)

    \(\frac{V_{1} \times 640}{\left ( 273 + 47 \right )} = \frac{620 \times 300}{\left ( 273 + 27 \right )}\)

    \(V_{1} = \frac{620 \times 300 \times 320}{640 \times 300} = 310 cc\)

The number of significant figures for the three numbers 161 cm, 0.161 cm, 0.0161 cm are

  • Solution

    We know that all non-zero digits are significant and the zeros at the beginning of a number are not significant.

    Therefore number 161 cm, 0.161 cm and 0.0161 cm have 3, 3 and 3 significant figures respectively.

In the final answer of the expression \(\frac{\left ( 29.2 – 20.2 \right )\left ( 1.79 \times 10^{5} \right )}{1.37}\) the number of significant figures is :

  • Solution

    On calculation we find

    \(\frac{\left ( 29.2 - 20.2 \right )\left ( 1.79 \times 10^{5} \right )}{1.37} = 1.17 \times 10^{6}\)

    As the least precise number contains 3 significant figures therefore, answers should also contains 3 significant figures.

The molecular weight of O2 and SO2 are 32 and 64 respectively. At 15°C and 150 mm Hg pressure, one litre of O2 contains ‘N’ molecules. The number of molecules in two litres of SO2 under the same conditions of temperature and pressure will be:

  • Solution

    According to Avogadro's law "equal volumes of all gases contain equal number of molecules under similar conditions of temperature and pressure".

    Thus if 1 L of one gas contains N molecules, 2 L of any other gas under the same conditions of temperature and pressure will contain 2N molecules.

Which one of the following pairs of compounds illustrate the law of multiple proportions ?

  • Solution

    SnCl2                                              SnCl4
    119 : 2 × 35.5                            119 : 4 × 35.5

    Chlorine ratio in both compounds is

    = 2 × 35.5 : 4 × 35.5 = 1 : 2

The weight of NaCl decomposed by 4.9g of H2SO4, if 6 g of sodium hydrogen sulphate and 1.825 g of HCl, were produced in the reaction is:

  • Solution

    According to law of conservation of mass "mass is neither created nor destroyed during a chemical change"

    ∴ Mass of the reactants = Mass of products

    x + 4.9 = 6+ 1.825

    or x = 2.925 g

1021 molecules are removed from 200 mg of CO2. The moles of CO2 left are :

  • Solution

    No. of moles = \(\frac{Wt. \; in \; g}{Mol. \; wt}\)

    No. of moles in 200 mg = \(\frac{200}{1000 \times 44}\)

    = 4.5 × 10-3 moles

    No. of moles in 1021 molecules

    = \(\frac{10^{21}}{6.02 \times 10^{23}} = 1.67 \times 10^{-3} moles\)

    No. of moles left = (4.5 – 1.67) × 10-3 = 2.88 × 10-3

0.4 moles of HCl and 0.2 moles of CaCl2 were dissolved in water to have 500 mL of solution, the molarity of Cl–ion is:

  • Solution

The empirical formula of a compound is CH2O. Its molecular weight is 180. The molecular formula of compound is :

  • Solution

    Empirical formula weight = 12 + 2 + 16 = 30

    \(n = \frac{180}{30} = 6\)

    Molecular formula = (CH2O)6 = C6H12O6

The oxide of an element contains 67.67% oxygen and the vapour density of its volatile chloride is 79. Equivalent weight of the element is:

  • Solution

    Equivalent weight of an element is its weight which reacts with 8 gm of oxygen to form oxide.

    Thus eq. weight of the given element \(\frac{32.33}{67.67} \times 8 = 3.82\)

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Some Basic Concepts of Chemistry
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