A solution of x g of an optically active sub | vedclass

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Question

(C) The formula for specific rotation $[\alpha]$ is given by: $[\alpha] = \frac{\alpha}{l 	imes c}$,where $\alpha$ is the observed rotation in degree

Vedclass · Accepted Answer

$1.25$

Vedclass · Answer

(C) The formula for specific rotation $[\alpha]$ is given by: $[\alpha] = \frac{\alpha}{l 	imes c}$,where $\alpha$ is the observed rotation in degrees,$l$ is the length of the tube in decimeters $(dm)$,and $c$ is the concentration in $g/mL$.
Given: $[\alpha] = +16^o$,$\alpha = +2^o$,$l = 10 \ cm = 1 \ dm$.
The concentration $c$ is given by $c = \frac{x \ g}{10 \ mL}$.
Substituting the values into the formula: $16 = \frac{2}{1 	imes (x/10)}$.
This simplifies to $16 = \frac{20}{x}$.
Solving for $x$: $x = \frac{20}{16} = 1.25 \ g$.

$A$ solution of $x \ g$ of an optically active substance dissolved in $10 \ mL$ of alcohol is placed in a polarimeter tube of $10 \ cm$ length. The observed rotation is $+2^o$. If the specific rotation is $+16^o$,what is the value of $x$ in $g$?

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