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Atoms And Molecules

Question
CBSEENSC9004981
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50 g of 10% lead nitrate is mixed with 50 g of 10% sodium chloride in a closed vessel. After the reaction has taken place, it was found that 6.83 g of lead chloride was precipitated. Besides, the reaction mixture contained 90 g water and sodium nitrate. Calculate the amount of sodium nitrate formed.

Solution

50 g of 10% lead nitrate means the solution contains 5 g lead nitrate and 45 g water. Similarly, 50 g of 10% sodium chloride means the solution contains 5 g sodium chloride and 45 g water.
Thus total content before reaction = 5+5+90 = 100 g
After reaction, amount of water = 90 g
Amount of precipitate = 6.83 g
Since according to law of conservation of mass, the total mass of reaction mixture = 100 g
So, Amount of sodium nitrate = 100 - 90 - 6.83
                                       = 3.17 g

Some More Questions From Atoms And Molecules Chapter

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