Question

90 ml of 21% (w/v) hno3 is completely neutralised by 112 ml of x% (w/v) of koh. The value of x is

Answer 1

Answer:

Step1: Apply the molarity formula to calculate the moles of HNO3.

HNO3 moles = 2M x .250 L = 0.50 moles.

Step2: Convert the moles of HNO3 into mass.

Mass = moles x molar mass = 0.50 moles x 63.01g/mole = 31.5 g HNO3.

Thus, 31.5 g HNO3 will be present in 100g x 31.5 g HNO3/70g = 45g solution.

Answer 2

Answer:

Explanation:

90 × 21/63×10 = 112 × x/56×10

X= 15

Hope it helps!!!!