Browse Questions

# The Freezing point of ether was lowered by $0.60^{\large\circ}$C on dissolving 2.0gm of phenol in 100g of ether.Calculate the molar mass of phenol and comment on the result.Given $K_f$(ether)=$5.12Kkg mol^{-1}$

$\begin{array}{1 1}170.7gmol^{-1}\\270.7gmol^{-1}\\370.7gmol^{-1}\\470.7gmol^{-1}\end{array}$

Let $M$ be the molar mass of phenol in ether.
We will have ,molality of solution =$\large\frac{2.0}{(100/1000)}$
Now using the expression $-\Delta T_f=K_fm$
We get,
$0.60K=(5.12Kkg mol^{-1})(\large\frac{20}{M}\frac{g}{kg})$
Which gives $M=\large\frac{5.12\times 20}{0.60}$$g mol^{-1}$
$\Rightarrow 170.7gmol^{-1}$
Since the calculated molar mass is about twice of its actual molar mass $(94gmol^{-1})$,it may be concluded that the phenol in ether is almost present in dimerized form
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8873980045318604, "perplexity": 7547.432085867844}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128320174.58/warc/CC-MAIN-20170623202724-20170623222724-00078.warc.gz"}