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(A) The formula for specific rotation $[\alpha]$ is given by: $[\alpha] = \frac{\alpha}{l 	imes c}$ Where: $\alpha$ is the observed rotation = $+10^{

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$(+) 100^{\circ}$

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(A) The formula for specific rotation $[\alpha]$ is given by: $[\alpha] = \frac{\alpha}{l 	imes c}$ Where: $\alpha$ is the observed rotation = $+10^{\circ}$ $l$ is the path length in $dm$ = $1 \ dm$ $c$ is the concentration in $g/mL$ = $\frac{1 \ g}{10 \ mL} = 0.1 \ g/mL$ Substituting the values: $[\alpha] = \frac{10}{1 	imes 0.1} = \frac{10}{0.1} = +100^{\circ}$ Therefore,the specific rotation is $+100^{\circ}$.

$A$ natural product was isolated in the laboratory and its observed rotation was $+10^{\circ}$. When measured in a $1 \ dm$ sample tube containing $1 \ g$ of compound in $10 \ mL$ of water. What is the specific rotation of this compound?

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