Number of HC moleaules present in 10 mLd0.1Nsolution is:(1) 6.022 x102

1.	Number of HC moleaules present in 10 mLd0.1Nsolution is:(1) 6.022 x1023(3) 6.022 x102143.(2) 6.023 x10246.022 x10
Answer» 4 is the correct option Molarity = moles of solute/liters of solution or, for our specific purpose here moles of solute (HCl) = liters of solution * Molarity 10 ml = 0.01 liters moles of solute (HCl) = (0.01 liters)(0.1 M HCl) = 0.001 moles of HCl hence molecule=0.0016.02310^23=6.02310^20

Number of HC moleaules present in 10 mLd0.1Nsolution is:(1) 6.022 x1023(3) 6.022 x102143.(2) 6.023 x10246.022 x10

Answer»

4 is the correct option

Molarity = moles of solute/liters of solution

or, for our specific purpose here

moles of solute (HCl) = liters of solution * Molarity

10 ml = 0.01 liters

moles of solute (HCl) = (0.01 liters)*(0.1 M HCl)

= 0.001 moles of HCl

hence molecule=0.001*6.023*10^23=6.023*10^20