Question

Concentration 1.00 M of NaOH Volume of Solution 2.5 mL added to solution 4.0 mL added to solution 2.00 M of Caci, 2.0 mL added to solution 0.75 M of K,POL Each of these three solutions are mixed into a beaker, find the molarity of each of the three individual solutions in the entire solution.

Answer 1

XI-3 SOLUTION Concentration of NaOH = 1.00M Volume of solution - 2.5 mL is 2. & Moles of NaOH taken ? = Concentration & volume = 1.00 2.5x 10-3 moles = 2.5 x 10-3 moles Similarly, Moles of Call, taken Concentration & volume 2.00×4.0×10-3 moles = 8 x 10-3 moles Moles of K₂PO4 taken = Concentration & volume Pa 0.75 x 2 x 10-3 moles = 1.5 x 10-3 moles to 2014 miature in ritstaand Total volume of mixture de - (2.5 & 4.0 & 2.0 JmL = 8.5 mL = 8.5 x 10-3 litres

In final solution , Molarity of NaOH - Moles Volume 2.5 x 10-3 8.5x10-3 3 moles / litre 0.29 M 8x10-3 moles (litre. Molarity of Calle - 8.5 x 10-3 0.94 M moles / litre Molarity of K₂ Pou - 1.5x 10-3 8.5 x 103 0.18 M
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