Question

11.00 mmol of of 1.00 M NaCl. If you transfer 50 g solute per er Ul Solution? 12. A beaker contains 100 g of 1.00 M NACH of the solution to another be solution remaining in the first beaker? he solution to another beaker, what is the molarity of the PROBLEMS

Answer 1

Suppose and 100 g solution - 100 ml solution 509 solution 350 ml_solution. 1009 or 100 mL of 1:00 M_Nach means that_molarity of solution is mole per litre 1 mole Nall mole Mall 1000 ml /1 litre contains - 1 ml solution contains and 100 me solution contains 000 1 X 100 1000 = 0.1 males of Nall. Similarly 50 ml solution of Nase contains -, L 150 1000 -0.05 moles of Nall. Now, calculate molarity when solution is 100-9/100ml_ and after, when Salution is 50ml/50g.. Molarity 1. S = meles volume in litre L. 100 ml = 0.100 L 0.100L - IM M1 similarly _MO Materity ? 0.050 = 40M [:: 50 ml = 0.050 L) Malarity of beaker is the solution remaining in same ie 1.oom the first Note: Molarity is an upon amount intensive property it of sample! does not depond