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Vedmedyk [2.9K]
3 years ago
7

How many grams of solute are present in 615 ml of 0.630 m kbr?

Chemistry
1 answer:
Tom [10]3 years ago
7 0
Molarity is the number of moles of solute in 1 L of solution
molarity of KBr solution is 0.630 M
this means that there there are 0.630 moles of KBr in 1 L solution 
then in 615 mL number of KBr moles are - 0.630 mol/L x 0.615 L = 0.387 mol
mass of KBr is - molar mass x number of moles 
KBr molar mass is 119 g/mol x 0.387 mol = 46.1 g
mass of KBr is 46.1 g
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              According to the law of mass equation

                                     

                                       K_{c} = \frac{[SO_{3} ][NO]}{[SO_{2}][NO_{2}  ]}

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So. number of moles of NO₂ at equilibrium added = 0.158

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