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A solution containing 28g of phosphorus ...

A solution containing `28`g of phosphorus in `315`g `CS_(2)(b.p.46.3^(@)C`boils at `47.98^(@)C`. If `k_(b)`for `CS_(2)`is `2.34`K kg `mol^(-1)`. The formula of phosphorus is (at .massof P=`31`).

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The correct Answer is:
4
`/_\T=(1000xxK_(b)^(')xxw)/(mxxW)`
`m_("exp")=123.8`
`(m_(N))/(m_("exp"))=1-alpha+(alpha)/n`
`:' alpha=1implies(m_(N))/(m_("exp"))=(alpha)/n`
`31/123.8=1/nimpliesn~~4`
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