(i) The density of the water at room temperature is 0.1 g/mL. How many molecules are there in a drop of water if its volume is 0.05 mL?

(ii) An alloy of iron (53.6%), nickel (45.8%) and manganese (0.6%) has a density of 8.17 g $c{{m}^{-3}}$. Calculate the number of Ni atoms present in the alloy of dimensions 10.0 cm x 20.0 cm x 15.0 cm.

# The density of the water at room temperature

**prasanna**#1

**prasanna**#2

(i) Volume of a drop of water = 0.05 mL

Mass of a drop of water = volume X density = (0.05 mL) X (1.0 g/mL) = 0.05 g l1/2l

Gram molecular mass of water =2x 1+ 16 = 18g

18 g of water = 1 mol

.-. 0.05 g of water = 1 mol X (0.05 g) = 0.0028 mol

1 mole of water contains molecules = 6.022 x ${{10}^{23}}$

0.0028 mole of water will contain molecules = 6.022 X ${{10}^{23}}$ X 0.0028= 1.68x ${{10}^{21}}$ molecules

(ii) Volume of the alloy = (10.0 cm) X (20.0 cm) X (15.0 cm)

= 3000 $C{{m}^{3}}$

Mass of the alloy = density X volume = (8.17 g $C{{m}^{-3}}$ )x (3000 $C{{m}^{3}}$) = 24510 g

Mass of Ni in the alloy = (24510 g )x45.8 /100 = 11225.6 g .

59 g Ni have atoms = 6.022 x ${{10}^{23}}$

Gram atomic mass of Ni = 59 g

11225.6 g of Ni have atoms = 6.022 X ${{10}^{23}}$ X 11225.6 g/59.0 g = 1.15x ${{10}^{26}}$ atoms