(C) The formula for specific rotation is $[\alpha] = \frac{\alpha}{l \times c}$,where $\alpha$ is the observed rotation in degrees,$l$ is the path len

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(C) The formula for specific rotation is $[\alpha] = \frac{\alpha}{l \times c}$,where $\alpha$ is the observed rotation in degrees,$l$ is the path length in decimeters $(dm)$,and $c$ is the concentration in $g/mL$. Given: $\alpha = -1.2^{\circ}$,$l = 5 \ cm = 0.5 \ dm$,and $c = 6.15 \ g / 100 \ mL = 0.0615 \ g/mL$. Substituting these values: $[\alpha] = \frac{-1.2}{0.5 \times 0.0615} = \frac{-1.2}{0.03075} = -39^{\circ}$.

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

The concentration of an organic compound in chloroform is $6.15 \ g$ per $100 \ mL$ of solution. $A$ portion of this solution in a $5 \ cm$ polarimeter tube causes an observed rotation of $-1.2^{\circ}$. What is the specific rotation of the compound?

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A
$+12^{\circ}$
B
$-3.9^{\circ}$
C
$-39^{\circ}$
D
$+61.5^{\circ}$

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The concentration of an organic compound in chloroform is $6.15 \ g$ per $100 \ mL$ of solution. $A$ portion of this solution in a $5 \ cm$ polarimeter tube causes an observed rotation of $-1.2^{\circ}$. What is the specific rotation of the compound?

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Three point charges $+q, -2q$ and $+q$ are placed at points $(x = 0, y = a, z = 0)$,$(x = 0, y = 0, z = 0)$ and $(x = a, y = 0, z = 0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are

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