the number of moles of KCl in 100ml of 3 molar solution is
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Given : Concentration of solution = 3M
No. of Mole in 0.1 lt of solutions (n) = Molarity x Volume =3*0.1=0.3moles
Therefore, no. of moles = 0.3
shaikmohaseen:
how is it 0.1m
No of Moles of HCl = 0.365/36.5 = 0.01
Volume = 0.1 l ; Molarity = n/V = 0.01/0.1 = 0.1
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calculate the weight of NaCl present in 250 g of 0.01 m
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However 0.01m Implies that 0.01moles of nacl in 1kg solvent. This Implies That 0.01*0
However 0.01m Implies that 0.01moles of nacl in 1kg solvent. This Implies That 0.01*0
This Implies That 0.01*0.25 moles of NaCl are present.
0.0025*(23+35.5)grams of NaCl is Present in the solution i.e., 0.14625g of NaCl is Present in 250g of 0.01m solution.
0.0025*(23+35.5)grams of NaCl is Present in the solution i.e., 0.14625g of NaCl is Present in 250g of 0.01m solution.
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