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How many atoms of each type present in 143 g of washing soda `(Na_(2)CO_(3).10H_(2)O)` ? |
Answer» Molar mass of washing soda `(Na_(2)CO_(3).10H_(2)O)=2xx23+12+3xx16+10xx18` `=46+12+4180=286` g 286 g of washing soda = 1 gram mol 143 of washing soda `= (1)/(286) xx 143 = 0.5` gram mol (i) No. of sodium atoms present 1 gram mole of washing soda contains Na atoms `= 2xx 6.022 xx 10^(23)` 0.5 gram mole of washing soda contains Na atoms `= 2xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(23)` (ii) No. of carbon atoms present 1 gram mole of washing soda contains C atoms `= 6.022 xx 10^(23)` `0.5` gram mole of washing soda contains C atoms `= 6.022 xx 10^(23)` 0.5 gram mole of washing soda contains C atoms `= 6.022 xx 10^(23) xx 0.5 = 3.011 xx 10^(23)` (iii) No. of oxygen atoms present 1 gram mole of washing soda contains O atoms `= 13 xx 6.022 xx 10^(23)` `0.5` gram mole of washing soda contains O atoms `= 13 xx 6.022 xx 10^(23) xx 0.5 = 3.914 xx 10^(24)` (iv) No. of hydrogen atoms present 1 gram mole of washing soda contains H atoms `= 20 xx 6.022 xx 10^(23)` `0.5` gram mole of washing soda contains H atom `= 20 xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(24)`. |
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