Answer:
Q.1
Given-
Volume of solution-1 L
Molarity of solution -6M
to find gms of AgNO3-?
Molarity = number of moles of solute/volume of solution in litre
number of moles of solute = 6×1= 6moles
one moles of AgNO3 weighs 169.87 g
so mass of 6 moles of AgNO3 = 169.87×6=1019.22
so you need 1019.22 g of AgNO3 to make 1.0 L of a 6.0 M solution
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0.125 g=(0.125 g)(1000 mg/1g)=125 mg.
Then, we need 125 mg of ampicillin.
5 ml of liquid suspension contains 250 mg of ampicilling , therefore:
5 ml----------------250 mg of ampicilling
x--------------------125 mg of ampicilling
x=(5 ml * 125 mg of ampicilling) / 250 mg of ampicilling=2.5 ml
Answer: we require 2.5 ml
