The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance B from the base line are 3.25 cm .2 .08 cm and 1.05 cm. respectively. The ratio of Rf values of A to B is

Question

The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance B from the base line are 3.25 cm .2 .08 cm and 1.05 cm. respectively. The ratio of Rf values of A to B is  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

ROH + CH 3 MgI arrow ROMgI + CH 4( g ) moles of CH 4= moles of ROH ⇒ ( V /22400)=( m / M . M ) (Assuming NTP Condition) ⇒ (3.1/22400)=(4.5 × 10-3/ M . M) ⇒ MM =32.51 Nearest Integer =33

Q. The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance $A$ and substance $B$ from the base line are $3.25 c m .2.08 c m$ and $1.05 c m$ . respectively. The ratio of $R_{f}$ values of $A$ to $B$ is _______

$RO H + C H_{3} M g I \to ROM g I + C H_{4} (g)$
moles of $C H_{4} =$ moles of $RO H$
$\Rightarrow \frac{V}{22400} = \frac{m}{M . M}$ (Assuming NTP Condition)
$\Rightarrow \frac{3.1}{22400} = \frac{4.5 \times 1 0 ^{- 3}}{M . M}$
$\Rightarrow MM = 32.51$
Nearest Integer $= 33$

Q. The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance A and substance B from the base line are 3.25cm.2.08cm and 1.05cm. respectively. The ratio of Rf​ values of A to B is _______

ROH+CH3​MgI→ROMgI+CH4​(g) moles of CH4​= moles of ROH ⇒22400V​=M.Mm​ (Assuming NTP Condition) ⇒224003.1​=M.M4.5×10−3​ ⇒MM=32.51 Nearest Integer =33

Q. The separation of two coloured substances was done by paper chromatography. The distances travelled by solvent front, substance $A$ and substance $B$ from the base line are $3.25 c m .2.08 c m$ and $1.05 c m$ . respectively. The ratio of $R_{f}$ values of $A$ to $B$ is _______

$RO H + C H_{3} M g I \to ROM g I + C H_{4} (g)$
moles of $C H_{4} =$ moles of $RO H$
$\Rightarrow \frac{V}{22400} = \frac{m}{M . M}$ (Assuming NTP Condition)
$\Rightarrow \frac{3.1}{22400} = \frac{4.5 \times 1 0 ^{- 3}}{M . M}$
$\Rightarrow MM = 32.51$
Nearest Integer $= 33$