A mixture of NaCl and NaBr has a mass of 2.07 g and is found to contain 0.73 g of Na. What is the mass of NaBr in the mixture?
no of mol of Na IN SAMPLE = 0.73/23 = 0.03174 mol
Mass of sample = mass of NaBr + mass of NaCl
2.07 = n*Mwt of NaBr + n*Mwt of NaCl
2.07 = (x*103+(0.03174-x)*58.5)
x = no of mol of NaBr in sample = 0.00479 mol
mass of NaBr in the mixture = 0.00479*103
= 0.4934 g
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