6. Calculate the molarity of 444 g KBr in 800 ml of solution. 31 7. How many mol

Question

6. Calculate the molarity of 444 g KBr in 800 ml of solution. 31 7. How many moles of KBr are contained in 110 ml of a 2.0 x10* mM KBr solution? 8. How would you prepare 500 mL of a 2 M KCN solution? 9. What concentration is produced after S ml of a 10 M stock KCN solution is diluted 100 mL? 10. After balancing the equation, write an ionic and net ionic equation for the reaction. 11. After balancing the equation, write an ionic and net ionic equation for the reaction 12. A 50.00-mL sample of an unknown HNO, solution requires titration with 8.2 mL of 2 M NaOH to reach the equivalence point. What is the concentration of the unknown HNOs solution? 13.Ildentifythe acid, bae, conjugate acid.and conjugate base in the followving reaction.

Explanation / Answer

Answer 6.

MW of KBr is 119

So, molarity of 1000 mL solution of 119 g of KBr is 1 (M )

then, 800 mL solution of 44.4 g of KBr is ( 44.4 x 800) / (1000 x 119) = 0.2985 (M)

Answer 7.

110 mL of 2.0 x 10-2 mM KBr solution = 110 mL of 2.0 x 10-5 M KBr solution

So, 1000 mL of  2.0 x 10-5 M KBr solution contain  2.0 x 10-5 mole KBr

Then, 110 mL of 2.0 x 10-5 M KBr solution contain  (2.0 x 10-5 x 110 ) /1000 mole KBr = 22 x 10-7 moles of KBr

Answer 8.

MW of KCN is 65.12

So, 1000 mL of 1M KCN solution contain 65.12 g of KCN

Then, 500 mL of 2M KCN solution contain (65.12 x 500 x 2) / 1000 g = 65.12 g of KCN

We have to dissolve 65.12 g of KCN in 500mL water to prepare 500 mL of 2M KCN solution.

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