Four values of normality for NaOH... I did this and I got it wrong. My initial v

Question

Four values of normality for NaOH...
I did this and I got it wrong. My initial values in ppth were 0.0988, 0.0998, 0.0998, and 0.0998. I did Q testing and 5ppth testing to eliminate 0.0988, and got it all wrong. Please help. Clark College Chemistry 153 4. values of the normality for a NaoH solution were found to be 0.09987 N, 0.0980 N, 0.09882 N and 0.09981 N. a.) Round all values to ppth precision, and place these rounded values in the boxes provided. b) Check to see if the "odd looking" value should be retained, using the Q test and the 5 ppth test. Circle any normality data that is rejected in the table below. Calculate the mean, standard deviation and RSD, and report them in the table below with proper significant figures. Concentration NaOH Trial 5ppth Mean: Standard Deviation: RSD (ppth):

Explanation / Answer

Average=0.09987+0.09980+0.09882+0.09981/4=0.3983/4=0.09956

Deviation1=0.09987-0.09956=0.00031

Deviation 2=0.09980-0.09956=0.00024

Deviation3= 0.09956-0.09882=0.00074

Deviation4=0.09981-0.09956=0.00025

Average deviation=(0.00031+0.00024+0.00025+0.00074)/4=0.000149/4=0.000037

parts per thousand=(0.000037/0.09956)*1000

ppth precision values are 0.0999, 0.0998, 0.0998, 0.0998

Mean=0.09956

Standard deviation=root[{(0.00031)^2+(0.00024)^2+(0.00074)^2+(0.00025)^2}/{4-1}]=root{(0.0000000961+0.0000000576+0.0000005476+0.0000000625)/3}=root(0.0000007638/3)=5.04*10^-4

RSD=100S/mean

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