Water percent

A 15 g sample of the substance contains 70% water. After drying, the weight was reduced to 9 g. What percentage of water is now in the sample?

Result

p =  50 %

Solution:

m1=15 g q=70%=70100=0.7 m2=9 g  v=q⋅ m1=0.7⋅ 15=212=10.5 g s=m1−v=15−10.5=92=4.5  p=100⋅ m2−sm2=100⋅ 9−4.59=50=50%m_{1}=15 \ \text{g} \ \\ q=70 \%=\dfrac{ 70 }{ 100 }=0.7 \ \\ m_{2}=9 \ \text{g} \ \\ \ \\ v=q \cdot \ m_{1}=0.7 \cdot \ 15=\dfrac{ 21 }{ 2 }=10.5 \ \text{g} \ \\ s=m_{1} - v=15 - 10.5=\dfrac{ 9 }{ 2 }=4.5 \ \\ \ \\ p=100 \cdot \ \dfrac{ m_{2}-s }{ m_{2} }=100 \cdot \ \dfrac{ 9-4.5 }{ 9 }=50=50 \%

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