Question

26) If 250 mL of 4,00 M CaBr, is diluted to 1.60 L. the resulting solution contains A) 0.400 mole of CaBr B) 1,00 mole of CaBr C) 2.40 moles of CaBr D) 100 mole of CaBr E) 0.0500 moles of CaBr 27) If 850 mlL of 1.36 M NH CI is diluted to 4.50 L, the concentration of the resulting solution is A) 0.0356 M B) 0.200 M C) 4.00 M D) 40.0 M E) 0.257 M 28) To prepare 25 mL of 0.25 M MgCl2 solution from a 0.75 M MgCl, stock solution, you should mix A) 10 mL of stock solution mixed with enough water to make 25 mL B) 12 mL of stock solution mixed with enough water to make 25 mL C) 17 mL of stock solution mixed with enough water to make 25 mL D) 9.6 mL of stock solution mixed with enough water to make 25 mL. E) 8.3 mL of stock solution mixed with enough water to make 25 mL 29) What volume, in liters, of 0.150 Ba(OH)2 solution is needed to react completely with 0.200 L of a 0.300 M HNO, solution according to the equation Ba(OH)2+2HNO, Ba(NO3) + 2H20 A) 0.150 L Ba(OH) B) 0.200 L Ba(OH) C) 0.300 L Ba(OH) D) 0.600 L Ba(OH)2 E) 0.250 L Ba(OH)

Answer 1

Solution :

26. Ans (B) 1.00 moles of CaBr₂

    4M CaBr₂ means that one litre contains 4 moles.

Therefore 250 ml of 4M CaBr2 contains 4 x 250/1000 = 1 mole.

When it is diluted to 1.6 L, this 1 mole will be present in 1.6 Litres.

27. Ans (E) 0.257 M

850 ml of 1.36 M NH4Cl contains 1.36 x 850 /1000 = 1.156 moles

When it is diluted to 4.5 L, this 1.156 moles will be present in 4.5 L

Therefore number of moles per L = 1.156/4.5 = 0.257 , that is the resulting solution has the concentration of 0.257M

28. Ans (E) 8.3 ml of stock solution mixed with enough water to make 25 ml.

25 ml of 0.25 M MgCl₂ contains 25 x 0.25/1000 = 0.00625 moles.

So to prepare 25 ml of 0.25 M MgCl_2, we need 0.00625 moles

This 0.00625 moles need to be obtained from the stock solution of 0.75 M MgCl₂

Suppose A ml of the stock solution contains 0.00625 moles , then A x 0.75/1000 = 0.00625

Therefore A = 0.00625 x 1000 / 0.75 = 8.3 ml.

29. Ans (B) 0.200 L Ba(OH)₂

As per equation 2 moles of HNO3 can react with 1 mole of Ba(OH)2

Therefore 0.300 M of HNO3 can react with 0.150 M of Ba(OH)2

So, 200 ml of 0.300 M HNO3 can react with 200 ml of 0.150M Ba(OH)2