JEE MAIN - Chemistry (2022 - 25th July Evening Shift - No. 21)
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 _____________.
Answer
2
Explanation
$$\mathrm{R}_{\mathrm{f}}=\frac{\text { Distance travelled by the substance }}{\text { Distance travelled by the solvent front }}$$
$$\left(R_{f}\right)_{A}=\frac{2.08}{3.25}$$
$$\left(R_{f}\right)_{B}=\frac{1.05}{3.25}$$
$$\frac{\left(R_{f}\right)_{A}}{\left(R_{f}\right)_{B}} \simeq 2$$
$$\left(R_{f}\right)_{A}=\frac{2.08}{3.25}$$
$$\left(R_{f}\right)_{B}=\frac{1.05}{3.25}$$
$$\frac{\left(R_{f}\right)_{A}}{\left(R_{f}\right)_{B}} \simeq 2$$
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