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Free CBSE - Class 11 Chemistry - Some Basic Concepts of Chemistry worksheets

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1.
Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:
Isotope Isotopic molar mass Abundance
$^{36}Ar$ 35.96755 g $mol^{-1}$ 0.337%
$^{38}Ar$ 37.96272 g $mol^{-1}$ 0.063%
$^{40}Ar$ 39.9624 g $mol^{-1}$ 99.600%
2.
Calculate the number of atoms in each of the following (i) 52 moles of Ar
3.
Calculate the number of atoms in each of the following (ii) 52 u of He
4.
Calculate the number of atoms in each of the following (iii) 52 g of He.
5.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula
6.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (ii) molar mass of the gas
7.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (iii) molecular formula.
8.
Calcium carbonate reacts with aqueous HCl to give $CaCl_2$ and $CO_2$ according to the reaction, $CaCO_3 (s) + 2 HCl (aq) \rightarrow CaCl_2 (aq) + CO_2(g) + H_2O(l)$. What mass of $CaCO_3$ is required to react completely with 25 mL of 0.75 M HCl?
9.
Chlorine is prepared in the laboratory by treating manganese dioxide ($MnO_2$) with aqueous hydrochloric acid according to the reaction
$4 HCl (aq) + MnO_2(s) \rightarrow 2H_2O (l) + MnCl_2(aq) + Cl_2 (g)$
How many grams of HCl react with 5.0 g of manganese dioxide?
10.
How many significant figures should be present in the answer of the following calculations? (iii) 0.0125 + 0.7864 + 0.0215
- OR -

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