At 25° C consider the density of water is 1 g/L and that of propanol to be 0.925 g/L . What volume of propanol will have same number of molecules as present in 210 mL of water?

Question

At 25° C consider the density of water is 1 g/L and that of propanol to be 0.925 g/L . What volume of propanol will have same number of molecules as present in 210 mL of water? (A)  757 mL  (B)  825 mL  (C)  646 mL  (D)  437 mL

Tardigrade Admin · Accepted Answer

textDensity =( textMass/ textVolume) For water: 1=(18/VH2 O) [∵ Molar mass of H2O=18 g mol-1] VH2 O=18 L ⇒ NA molecules For propanol: 0.925=(60/V textpropanol) [∵ Molar mass of CH3CH2CH2 OH=60 g mol-1] V textpropanol=64.86 L ⇒ NA molecules Ratio: (V textpropanol/VH2 O) =(64.86/18) V textpropanol=3.603× VH2 O ⇒ V textpropanol=3.603×210 =756.63 =757 mL

Q. At $2 5^{\circ} C$ consider the density of water is $1 g / L$ and that of propanol to be $0.925 g / L$ . What volume of propanol will have same number of molecules as present in $210 m L$ of water?

$757 m L$

$825 m L$

$646 m L$

$437 m L$

Q. At 25∘C consider the density of water is 1g/L and that of propanol to be 0.925g/L . What volume of propanol will have same number of molecules as present in 210mL of water?

757mL

825mL

646mL

437mL

Q. At $2 5^{\circ} C$ consider the density of water is $1 g / L$ and that of propanol to be $0.925 g / L$ . What volume of propanol will have same number of molecules as present in $210 m L$ of water?

$757 m L$

$825 m L$

$646 m L$

$437 m L$